1. Class 10: Introduction to Linear Programming
OPSM301 Spring 2012
Class 10:
Introduction to Linear Programming

2. Question-a
We can produce 2 products: Hockey Sticks (H) and Chess Sets (C)
Marginal profit per product: H:$3, C:4$ per unit
H requires 4 hours of processing at machine center A and 2 hours at machine center B
C requires 5 hours at machine center A, 3 hours at machine center B
Assume we have 120 hrs capacity available on A, 120 hrs on B
How many H and C should be produced per day to maximize profit?
Machine A
Machine B
Chess Set
Hockey Stick
C: hr/unit hr/unit
H: hr/unit hr/unit
Notice that for both products, the bottleneck is Machine A
Profit per minute on A: Chess Sets:20 units/day x 4 $/unit = 80 $/day
Hockey Stick: 30 units/day x 3 $/unit = 90 $/day

3. Question-b Machine A Machine B Chess Set Hockey Stick
We can produce 2 products: Hockey Sticks (H) and Chess Sets (C)
Marginal profit per product: H:$3, C:4$ per unit
H requires 4 hours of processing at machine center A and 2 hours at machine center B
C requires 3 hours at machine center A, 4 hours at machine center B
Assume we have 120 hrs capacity available on A, 120 hrs on B
How many H and C should be produced per day to maximize profit?
Machine A
Machine B
Chess Set
Hockey Stick
C: hr/unit hr/unit
H: hr/unit hr/unit
Notice that for Chess Sets, Machine B is bottleneck,
for hockey sticks, Machine A is bottleneck

4. What is the production plan to maximize profit?
C is more profitable. Produce only C?
C: Bottleneck is B, capacity=120/4=30/day
Produce only C  30x4=120$/day
H: Bottleneck is A, capacity=30/day.
Produce only H  30x3=90$/day
Solution 1: 0 H, 30 C
120$
Solution 2: 30 H, 0 C
90$
Can we do better?
Best solution:produce 12 H, 24 C:
Both machines are utilized for 120 hrs
Profit=(12x3) + (24x4)=132$/day
Solution 3: 12 H, 24 C
132$
This is a very simple product-mix problem.
Real life problems are more complicated. How can we solve such problems?
 Linear Programming

5. Linear Programming (LP)
Making decisions on the allocation of scarce resources in environments of certainty using a mathematical optimization approach

6. History of LP
Dantzig devised the method to solve LPs in 1947, working for US Air Force, for the training and logictics planning
First large scale application for the diet problem for soldiers (the cheapest adequate diet)
First commercial applications in 1952, for optimal blending of petroleun products to make gasoline
Typical Applications:
Develop a production plan that minimizes production and inventory costs while meeting demand
Construct a portfolio of securities that maximizes return while keeping "risk" below a predetermined level
Develop an advertising strategy to maximize exposure of potential customers while staying within a predetermined budget

7. Conditions for Applicability of Linear Programming
Resources must be limited
There must be an objective function
There must be linearity in the constraints and in the objective function
Resources and products must be homogeneous (e.g. All hockey sticks are the same)
Decision variables must be divisible (continuous variables) and non-negative

8. Components of Linear Programming
A specified objective or a single goal, such as the maximization of profit, minimization of machine idle time etc.
Decision variables represent choices available to the decision maker in terms of amounts of either inputs or outputs
Constraints are limitations which restrict the alternatives available to decision makers

9. Components of Linear Programming
There are three types of constraints:
(=<) An upper limit on the amount of some scarce resource
(>=) A lower bound that must be achieved in the final solution
(=) An exact specification of what a decision variable should be equal to in the final solution
Parameters are fixed and given values which determine the relationships between the decision variables of the problem

10. Formulating our Product-mix Example as an LP
Decision variables: H:number hockey sticks C:number chess sets Objective Function: Maximize Z = $3H + $4C Constraints: Subject to 4H + 3C < 120 (machine center A constraint) 2H + 4C < 120 (machine center B constraint) H, C > 0 (non-negativity constraints)

11. Solving Linear Programs
QMBU 302 (an operations management elective) focuses on the solution of optimization problems
Linear programs can be solved using Excel Solver
We will learn using basic Solver next Thursdsay, March 15 (in SOS180)

12. Excel Solver Solution Optimal Solution Hockey Sticks Chess Sets Total
Hockey Sticks
Chess Sets
Total
changing cells 12 24
profit 3 4
132
Resource Used
 Capacity
Machine A
120
Machine B 2
Optimal Solution

13. Example: Giapetto's Woodcarving
Giapetto's Woodcarving, Inc., manufactures two types of wooden toys: soldiers and trains. A soldier sells for $27 and uses $10 worth of raw materials. Each soldier that is manufactured increases Giapetto's variable labor and overhead costs by $14. A train sells for $21 and uses $9 worth of raw materials. Each train built increases Giapetto's variable labor and overhead costs by $10. The manufacture of wooden soldiers and trains requires two types of skilled labor: Carpentry and finishing. A soldier requires 2 hours of finishing labor andone hour of carpentry labor. A train requires 1 hour of finishing labor and 1 hour of carpentry labor. Each week, Giapetto can obtain all the needed raw material but only 100 finishing hours and 80 carpentry hours. Demand for trains is unlimited, but at most 40 soldiers are sold each week. Giapetto wishes to maximize weekly profit. Formulate a mathematical model of Giapetto's situation that can be used to maximize Giapetto's weekly profit.

14. Towards the Mathematical Model:
Define (decision variables)
x1 : number of soldiers produced each week
x2 : number of trains produced each week
Objective function:
maximize weekly profit = weekly profit from soldiers + weekly profit from trains
Constraints:
each week, no more than 100 hours of finishing time may be used
each week, no more than 80 hours of carpentry time may be used
each week, the number of soldiers produced should not exceed 40 because of limited demand

15. The Linear Programming Model:
max z = 3x1 + 2x2 subject to 2x1 + x2  100 (finishing hours) x1 + x2  80 (carpentry hours) x1  40 (demand for soldiers) x1  0 (nonnegativity constraint) x2  0 (nonnegativity constraint)

16. The Excel Model soldiers trains Total(objective) changing cells 20 60
soldiers
trains
Total(objective)
changing cells 20 60
profit 3 2
180
used
capacity
finishing 1
100
carpenter 80
demand 40
Filled in by Excel Solver

17. Reading the variable information
The optimal solution for Giapetto is to produce 20 soldiers and 60 trains per week, resulting in an optimal profit of $180. (The maximum possible profit attainable is $180, which can be achieved by producing 20 soldiers and 60 trains)

18. Common Application Areas
Product planning-product mix, diet problems
Manpower planning
Investment planning
Aggregate production planning
Transportation problem
Process control (cutting sheets etc.)
Distribution/routing
Plant location
Service productivity analysis (DEA)

19. Example Problem (old Exam Question)
The Huntz Company purchases cucumbers and makes two kinds of pickles: sweet and dill. The company policy is that at least 30%, but no more than 60%, of the pickles be sweet. The demand for pickles is
SWEET:5000 jars + additional 3 jars for each $1 spent on advertising
DILL:4000 jars + additional 5 jars for each $1 spent on advertising
Sweet and dill pickles are advertised separately.  The production costs are:  
SWEET:0.60 $/jar DILL:0.85 $/jar
and the selling prices are:  
SWEET:1.45 $/jar DILL:1.75 $/jar
Huntz has $16,000 to spend on producing and advertising pickles. Formulate an appropriate Linear Program.

20. Solution: LP Formulation
Xs: Number of Sweet pickle jars produced.
Xd: Number of Dill pickle jars produced.
As: Amount of advertisement done for Sweet Pickles in dollars
Ad: Amount of advertisement done for Dill Pickles in dollars
Objective; maximize profits: (Revenue - Cost)
max[(1,45* Xs + 1,75* Xd)-( 0,6* Xs + 0,85* Xd + As + Ad)]
Subject to:
Demand Constraints:
Xs ≤ 5, As
Xd ≤ 4, Ad
Budget Constraint:
0,6* Xs + 0,85* Xd + As + Ad ≤ 16,000
Ratio Constraint:
Xs / (Xs + Xd) ≥ 0,3  0,7 Xs – 0,3 Xd ≥ 0
Xs / (Xs + Xd) ≤ 0,6  0,4 Xs – 0,6 Xd ≤ 0
Non-negativity Constraints:
Xs, Xd, As, Ad ≥0

21. Next Time
More examples for LP models